loup blanc wrote:

$f:x\in\mathbb{R}^n \setminus \{\vec{0}\}\mapsto S_{n-1}\subset \mathbb{R}^n$ .
Then $im(Df_x)\subset T_{f(x)}(S_{n-1})$ , the tangent space to $S_{n-1}$ in the point $f(x)$ .
On the other hand, $f_{|{S_{n-1}}}=id$ . Thus $im(Df_x))=T_{f(x)}(S_{n-1})$ and $rank(Df_x)=n-1$ .
Thus $dim(\ker(Df_x))=1$ ; more precisely, since $f(x+\lambda x)=f(x)$ , we deduce that $\ker(Df_x)=span[x]$ .

Thank you very much for your response. I will write your solution in a way that is clear to me for my own reference.

Observe that $f(\vec{x}) = \frac{\vec{x}}{||\vec{x}||}: \mathbb{R}^n \setminus \{\vec{0}\} \to S_{n - 1}$ , so its derivative is a map between the tangent spaces of $\vec{x}$ and $f(\vec{x})$ , implying $Df_{\vec{x}}(T_{\vec{x}}(\mathbb{R}^n \setminus \{\vec{0}\})) \subseteq T_{f(\vec{x})}S_{n - 1}$ . We can see the two sets are equal as follows. For any $r > 0$ and $||\vec{x}|| = r$ , consider $f(\vec{x}) = \vec{x}/r$ as a map from $rS_{n - 1} \to S_{n - 1}$ . Differentiating gives $Df_{\vec{x}} = \frac{I}{r}: T_{\vec{x}}(rS_{n - 1}) \to T_{f(\vec{x})}S_{n - 1}$ . Both tangent spaces are isomorphic to $\mathbb{R}^{n - 1}$ because $rS_{n - 1}$ and $S_{n - 1}$ are $(n - 1)$ -dimensional, and since $\frac{I}{r}: \mathbb{R}^{n - 1} \to \mathbb{R}^{n - 1}$ is surjective, we have $Df_{\vec{x}}(T_{\vec{x}}(rS_{n - 1})) = T_{f(\vec{x})}S_{n - 1}$ . The left side is in $Df_{\vec{x}}(T_{\vec{x}}(\mathbb{R}^n \setminus \{\vec{0}\}))$ so we must have $Df_{\vec{x}}(T_{\vec{x}}(\mathbb{R}^n \setminus \{\vec{0}\})) = T_{f(\vec{x})}S_{n - 1}$ as well. Equivalently, $\text{rank}(Df_{\vec{x}}) = n - 1$ . Since $f(\lambda \vec{x}) = f(\vec{x})$ for any $\lambda \neq 0$ , we know $f$ is constant along $\mathbb{R}_{\neq 0}\vec{x}$ so zero is the result when its derivative is multiplied by any vector on that line. This implies $\mathbb{R}\vec{x} \subseteq \text{ker}(Df_{\vec{x}})$ . By the rank-nullity theorem, $\text{ker}(Df_{\vec{x}})$ has dimension $n - \text{dim}(\text{rank}(Df_{\vec{x}})) = 1$ , so we must have equality.

Again, much appreciate your response.